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To solve part (a) of this problem, we need to find the area of the shaded region \( R \), which is bounded by the curve \( x = (y - 3)^2 - 9 \), the x-axis, and the vertical line \( x = -9 \).

### Step-by-Step Solution:

1. **Equation of the curve**:
   \[
   x = (y - 3)^2 - 9
   \]
   This is a parabolic equation in terms of \( x \) as a function of \( y \). We need to determine the limits of integration along the \( y \)-axis.

2. **Find the intersection points**:
   The region is bounded by the line \( x = -9 \) and the curve. To find the limits of integration, set \( x = -9 \) in the equation of the curve:
   \[
   -9 = (y - 3)^2 - 9
   \]
   Simplifying:
   \[
   0 = (y - 3)^2
   \]
   This gives \( y = 3 \). So, the curve intersects the line \( x = -9 \) at the point \( (-9, 3) \).

   The other boundary is the x-axis, which is \( y = 0 \).

3. **Set up the integral**:
   The area can be found by integrating the difference between the curve \( x = (y - 3)^2 - 9 \) and the vertical line \( x = -9 \), from \( y = 0 \) to \( y = 3 \):
   \[
   \text{Area} = \int_0^3 \left[ (y - 3)^2 - 9 - (-9) \right] \, dy
   \]
   Simplifying the integrand:
   \[
   \text{Area} = \int_0^3 (y - 3)^2 \, dy
   \]

4. **Evaluate the integral**:
   First, expand \( (y - 3)^2 \):
   \[
   (y - 3)^2 = y^2 - 6y + 9
   \]
   So the integral becomes:
   \[
   \text{Area} = \int_0^3 (y^2 - 6y + 9) \, dy
   \]
   Now, integrate term by term:
   \[
   \int_0^3 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^3 = \frac{27}{3} = 9
   \]
   \[
   \int_0^3 -6y \, dy = \left[ -3y^2 \right]_0^3 = -3(9) = -27
   \]
   \[
   \int_0^3 9 \, dy = \left[ 9y \right]_0^3 = 27
   \]

   So, adding these results:
   \[
   \text{Area} = 9 - 27 + 27 = 9
   \]

Thus, the area of the shaded region \( R \) is **9 square units**.

---

Would you like more details or explanations on any part of the solution? 

Here are 5 related questions for further exploration:
1. How would you calculate the area if the curve was \( x = (y - 2)^2 - 4 \)?
2. What if the boundary was set at \( x = -10 \) instead of \( x = -9 \)?
3. How does the shape of the region change if the equation of the curve changes to \( x = (y - 4)^2 - 9 \)?
4. Can you find the centroid of the shaded region \( R \)?
5. How would the problem differ if you were calculating the volume of a solid generated by rotating this region around the \( y \)-axis?

**Tip**: When solving area problems bounded by curves, ensure to properly determine the limits of integration by finding the intersection points.

Find the area of the shaded region R bounded by the curve x = (y - 3)^2 - 9, the x-axis, and the vertical line x = -9.

Learn how to calculate the area of a region bounded by the curve x = (y - 3)^2 - 9, the x-axis, and the vertical line x = -9 using integral calculus. Detailed solution provided with step-by-step integration.

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